Answer
See graph and explanations.
Work Step by Step
Step 1. With $y=1+cos(x)$, we have $y'=-sin(x)$ which give the slope of tangent lines.
Step 2. At $x=-\pi/3$, $y=1+cos(-\pi/3)=3/2$ and $m_1=y'=-sin(-\pi/3)=\sqrt 3/2$, tangent line equation: $y-3/2=\sqrt 3/2(x+\pi/3)$ or $y=\sqrt 3x/2+\sqrt 3\pi/6+3/2$
Step 3. At $x=3\pi/2$, $y=1+cos(3\pi/2)=1$ and $m_2=y'=-sin(3\pi/2)=1$, tangent line equation: $y-1=x-3\pi/2$ or $y=x-3\pi/2+1$
